You haven't provided the coordinates of C and D, therefore, I cannot provide an exact solution. However, I'll tell you how to solve this problem and you can apply on the coordinates you have.
The general form of the linear equation is:y = mx + c
where:
m is the slope and c is the y-intercept
1- getting the slope:We will start by getting the slope of CD using the formula:
slope = (y2-y1) / (x2-x1)
We know that the line we are looking for is perpendicular to CD. This meas that the product of their slopes is -1. Knowing this, and having calculated the slope of CD, we can simply get the slope of our line
2- getting the y-intercept:To get the y-intercept, we will need a point that belongs to the line.
We know that our line passes through the midpoint of CD.
Therefore, we will first need to get the midpoint:
midpoint = (

)
Now, we will use this point along with the slope we have to substitute in the general equation and solve for c.
By this, we would have our equation in the form of:y = mx + c
Hope this helps :)
The answer is 75! Explanation: Use the distance formula:)
Answer:
pattern
Step-by-step explanation:
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The Volume of the given solid using polar coordinate is:![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2860%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
V= ![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2860%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
<h3>
What is Volume of Solid in polar coordinates?</h3>
To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.
Consider the cylinder,
and the ellipsoid, 
In polar coordinates, we know that

So, the ellipsoid gives

4(
) +
= 64
= 64- 4(
)
z=± 
So, the volume of the solid is given by:
V= ![\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B2%5Cpi%7D_%200%20%5Cint%5Climits%5E1_0%7B%7D%20%5C%2C%20%5B%5Csqrt%7B64-4r%5E%7B2%7D%20%7D-%20%28-%5Csqrt%7B64-4r%5E%7B2%7D%20%7D%29%5D%20r%20dr%20d%5Ctheta)
= 
To solve the integral take,
= t
dt= -8rdr
rdr = 
So, the integral
become
=
= 
=
so on applying the limit, the volume becomes
V= 
=![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2864-4%281%29%5E%7B2%7D%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864-4%282%29%5E%7B0%7D%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
V = ![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2860%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
Since, further the integral isn't having any term of
.
we will end here.
The Volume of the given solid using polar coordinate is:![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2860%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
Learn more about Volume in polar coordinate here:
brainly.com/question/25172004
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